GROUPS 2

Example 3 :

Let G = {x element of G|x is not equal to 1}. On G, define * by a*b = a+b-ab. Determine whether the set G is a group under *.

G1 : (To show associativity of *)
Let a, b, c element of G & a,b, c is not equal to 1.

(a*b)*c = a*(b*c)


LHS :

(a*b)*c
= (a+b-ab)*c let y = a+b-ab
= y*c
= y+c-yc
= a+b-ab+c-(a+b-ab)c
= a+b-ab+c-(ac+ab-abc)
= a+b-ab+c-ac-bc+(ab)c

RHS :

a*(b*c)
= a*y
= a+y-ay
= a+b+c-bc-a(b+c-bc)
= a+b+c-bc-ab-ac+a(bc)

Since + is commutative on Q & . is associative on Q.
Therefore, * is associative on G.

G2 : (To show that, a*e = a = e*a)

LHS :

a*e = a
a+e-ae = a
a+e(1-a) = a
e(1-a) = 0
e = 0

RHS :

e*a = a
e+a-ea = a
e(1-a) = 0
e = 0

Since, LHS = RHS for e = 0, 0 is identity element for binary structure (G,*).

G3 : (To find the inverse element)

a*a' = a'*a = e

Let a element of Q & a is not equal to 1.

LHS :

a*a' = 0
a+a'-aa' = 0
a'(1-a) = -a
a'= -a/1-a

RHS :

a'*a = 0
a'+a-a'a = 0
a'(1-a) = -a
a' = a/1-a

Since, LHS = RHS.
Then, a' = a/1-a is the inverse element of (G,*).
Thefore, (G,*) is a group.